Will the power dissipated in a resistor in an AC circuit vary with frequency? In a recent examination I had a question where the examiner asked if the brightness of a bulb connected to an AC source would vary with frequency of AC source. The right answer was that it would vary, as the frequency increased so would the brightness. I intuitively feel that it wouldn't. I couldn't find a relation between frequency and power dissipated by a quick web search(all the relations are in terms of voltage). How does the power vary with frequency ? In the question the circuit only has a source and a resistor.
 A: 
How does the power vary with frequency ?

For an ideal source and ideal resistor it doesn't vary.
For a real source and real resistor, connected by real wires, it will vary in a complex way that depends just exactly what is non-ideal about those components.

The right answer was that it would vary, as the frequency increased so would the brightness.

I can think of several reasons why the brightness would decrease as the frequency increases:


*

*At high frequency, the inductance of the connecting wires has higher impedance, reducing the voltage delivered to the bulb.

*At still higher frequency, the wires between the source and bulb form a loop antenna and radiate power away rather than deliver it to the bulb.

*At still higher frequency, the skin effect increases the effective resistance of the filament, reducing the power it consumes according to $P=V^2/R$.

*At still higher frequency, the skin effect increases the effective resistance of the interconnecting wires, reducing the voltage delivered to the bulb.
It's possible depending on the exact geometry of the components, that the order that these effects become significant is different than I suggested. It's also possible there's a resonance effect between the wire and bulb in some frequency band that results in high power delivered to the bulb. But these would depend on knowing the exact dimensions of the wires, bulb, etc., and couldn't be counted on in general to increase the bulb brightness with increasing frequency.
I can't think of any mechanism that would lead to the bulb getting brighter as frequency increases for arbitrary geometries, but I suppose there is likely something your instructor had in mind when they wrote the question. The mechanism might be hinted at in the specific wording of the question, if you read it carefully.
A: In practice (say, in engineering terms) the power dissipated in the resistive load is not a function of frequency. Electrical measurement tools use this fact to measure power independent of frequency in the following way: 
A non-inductive resistor is housed in a well-insulated container with known thermal properties. An electrical signal consisting of a mixture of unknown frequencies is applied to this load. A sensitive thermometer measures the rate of temperature rise inside the container and from this, the power being dissipated in the load can be back-calculated. This "calorimeter" technique overcomes any tendency on the part of ordinary power meters (i.e., meters that measure voltage and current) to have non-flat frequency response curves due to their own inductance and capacitance, which create errors in their measurements. 
A: A light bulb when operated from an variable frequency AC source will likely have reactive components, i.e. inductance and capacitance, which will have a resonant frequency. As the frequency of the source signal tends towards the resonant frequency of the bulb and wires the power delivered will vary. We would have to look at the maths to see how this could work.
